Дискретка.Лекции, литература / Chast1_2012

26.Пользуясь принципом двойственности, доказать самодвойственность функции, заданной формулой.

26.1.x2 # (x3 x1)x2

26.2.(x2 j x1 ! x2) x3

26.3.x2 ((x3 # x1) j x1)

26.4.x2 j (x3 ! (x1 ! x2))

26.5.(x2 ! (x2 _ x3)) x1

26.6.x1 (x2 # (x2 j x3))

26.7.x2((x1 ! x3) ! x2)

26.8.x1 _ ((x1 ! x2) # x3)

26.9.((x2 # x1) j x2)x3

26.10.x2 (x1 # (x3 j x1))

26.11.((x3 x2) ! x1) j x1

26.12.x1 ((x3 j x2) # x3)

26.13.(x3 j (x2 x1)) ! x3

26.14.(x1 # (x1 j x2)) # x3

26.15.x1 j (x3 ! (x3 _ x2))

26.16.x1 (x2 # (x3 j x2))

26.17.(x2 _ (x1 ! x3)) j x3

26.18.x2 # (x2 ! x1)x3

26.19.x2 _ (x3 # x1)x3

26.20.x2(x1 _ x2 j x3)

26.21.((x2 x3) ! x1) j x1

26.22.x1(x1 # x2) x3

26.23.x3 # (x2 # (x3 j x1))

26.24.((x3 # x1) ! x2)x1

26.25.x3 j ((x1 _ x2) ! x3)

26.26.(x1 j x2 ! x3)x3

26.27.(x2 # (x3 j x1)) _ x3

26.28.(x3 ! (x3 _ x1)) j x2

26.29.x2 # (x3 ! x1) j x1

26.30.((x2 j x1) # x1) x3

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27.Проверить самодвойственность функции f, заданной векторно.

27.1.f = (1011100001010101)

27.2.f = (1100010101100011)

27.3.f = (0100001110101110)

27.4.f = (0101111001011000)

27.5.f = (0001111010101001)

27.6.f = (1011100001001101)

27.7.f = (1010010110011010)

27.8.f = (1001010010010111)

27.9.f = (1011101010010010)

27.10.f = (0101011001010101)

27.11.f = (1000000011011111)

27.12.f = (0111001110101000)

27.13.f = (0000110101101110)

27.14.f = (1001101111010000)

27.15.f = (0110001110011001)

27.16.f = (0110101000110101)

27.17.f = (1101000101000111)

27.18.f = (0100101111010100)

27.19.f = (0100110110001101)

27.20.f = (0010110111100001)

27.21.f = (0101000011011101)

27.22.f = (0001010111010110)

27.23.f = (1110000110111000)

27.24.f = (0100101010001111)

27.25.f = (1110001100101010)

27.26.f = (1110000111001010)

27.27.f = (0110010101010101)

27.28.f = (1011101100100001)

27.29.f = (1011001010101001)

27.30.f = (1111000100110010)

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28.Подсчитать, сколькими способами можно заменить прочерки в вектореконстантами 0 и 1 так, чтобы получился вектор значений некоторой

самодвойственной функции.

28.1.= ( 10 0 )

28.2.= ( 0 01 0)

28.3.= (11 1 0 )

28.4.= ( 0 111 )

28.5.= ( 0 1 1 )

28.6.= ( 1 100 )

28.7.= (0 1 01 )

28.8.= ( 1 00 )

28.9.= ( 1 1 0 1)

28.10.= ( 1 1 00 )

28.11.= ( 0 0 0 )

28.12.= ( 110 0)

28.13.= ( 0 0 01)

28.14.= ( 1001 )

28.15.= ( 1 0 0 )

28.16.= (1 0 0 1 )

28.17.= (0 0 0 1)

28.18.= (0 0 10 )

28.19.= ( 01 0)

28.20.= ( 1 00 0 )

28.21.= ( 1 10 0 )

28.22.= ( 00 1 0 )

28.23.= (1 0 11 )

28.24.= ( 0 0 00 )

28.25.= ( 0 0 11 )

28.26.= ( 0011 )

28.27.= (0 01 0 )

28.28.= ( 1 1 0 )

28.29.= ( 1 01)

28.30.= (0 1 0 0 )

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29.Используя лемму о несамодвойственной функции, указатü, какие из переменных x1; x2; x3; x4 нужно заменить на x, а какие на x с тем, чтобы получить из функции f константу.

29.1.f(x1; x2; x3; x4) = (0011000001110011)

29.2.f(x1; x2; x3; x4) = (0101110001000101)

29.3.f(x1; x2; x3; x4) = (0001100101000111)

29.4.f(x1; x2; x3; x4) = (1010000101101010)

29.5.f(x1; x2; x3; x4) = (0110000101011001)

29.6.f(x1; x2; x3; x4) = (1101110011010100)

29.7.f(x1; x2; x3; x4) = (0101000001110101)

29.8.f(x1; x2; x3; x4) = (0100000101101101)

29.9.f(x1; x2; x3; x4) = (0101100101110101)

29.10.f(x1; x2; x3; x4) = (1111001010010000)

29.11.f(x1; x2; x3; x4) = (0110110011101001)

29.12.f(x1; x2; x3; x4) = (1000111000001110)

29.13.f(x1; x2; x3; x4) = (1000100100101110)

29.14.f(x1; x2; x3; x4) = (1011111010010010)

29.15.f(x1; x2; x3; x4) = (0111000011100001)

29.16.f(x1; x2; x3; x4) = (1010010001011010)

29.17.f(x1; x2; x3; x4) = (1001110100000110)

29.18.f(x1; x2; x3; x4) = (0001110011100111)

29.19.f(x1; x2; x3; x4) = (1001111100100110)

29.20.f(x1; x2; x3; x4) = (0010110100001011)

29.21.f(x1; x2; x3; x4) = (0101101110100101)

29.22.f(x1; x2; x3; x4) = (0100011100001101)

29.23.f(x1; x2; x3; x4) = (1101111011000100)

29.24.f(x1; x2; x3; x4) = (1010011100111010)

29.25.f(x1; x2; x3; x4) = (0111101101100001)

29.26.f(x1; x2; x3; x4) = (0001110011010111)

29.27.f(x1; x2; x3; x4) = (0000101000101111)

29.28.f(x1; x2; x3; x4) = (0011110010000011)

29.29.f(x1; x2; x3; x4) = (1001010001010110)

29.30.f(x1; x2; x3; x4) = (1101110111000100)

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30.Проверить, является ли функция линейной.

30.1.((x3 x1) ! (x2 j x1))(x3 x2)x3

30.2.(x1 (x1 x3)) ! (x2 x3 (x3 x1))

30.3.(x2 (x3x2)) # x1

30.4.x1 ! (x2 j (x2 _ x3))

30.5.(x1 j (x3 _ x2)) # x1

30.6.(x2 # x3)((x1 x2) # x2)

30.7.((x2 x3)x1) (x2 # x1)

30.8.((x2 x3) j x3) j x1

30.9.(x2 ! x1)(x1 j x2)x3

30.10.(x2 j x1) (x3 x1) (x1 _ x3)

30.11.(x3 (x2x3)) ((x2 _ x1)x1)

30.12.(x1 # x2) ! (x2 _ x3)

30.13.((x1 ! x3) ! (x2x3))

30.14.((x3 x1) j x1) # (x3 j x2)

30.15.x1 _ (x3 (x2 j x1))

30.16.(x1 ! x3) ! (x2 x3)

30.17.(x2 ! x3) j (x1 j x2)

30.18.(x2(x2 x3)) x3 (x1 j x2)

30.19.(x2x3 x2 x1) (x2 _ x3)

30.20.(x3 ! x2) j (x3 x1)

30.21.(x3 _ x1) (x2 # x3)

30.22.(x3 j x1) j (x1 # x2)

30.23.(x3 x3) j (x3 j x2)

30.24.(x1 (x2 # x3)) (x2 _ x3)

30.25.(x2 _ x1)(x3 ! x2)

30.26.((x1 x2) j x3) (x2 ! x3)

30.27.(x3 (x2x3)) j (x2 j (x2 x1))

30.28.((x1 _ x3) x2) j (x2x3 x3)

30.29.(x1 ! x2) _ (x1 (x2 x3))

30.30.(x1 # x3) (x3 (x1 _ x2))

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31.Проверить линейность функции f, заданной векторно.

31.1.f = (1110100011100100)

31.2.f = (1101001010110010)

31.3.f = (1101000011110001)

31.4.f = (0101110010011001)

31.5.f = (1010110000101011)

31.6.f = (1001111000011100)

31.7.f = (0100001100111011)

31.8.f = (0100001010111110)

31.9.f = (1100101010010110)

31.10.f = (1101001101000011)

31.11.f = (0110110010101010)

31.12.f = (1110000010110011)

31.13.f = (0111010110100100)

31.14.f = (0011010011111000)

31.15.f = (0001111000011011)

31.16.f = (1011001010100110)

31.17.f = (0001011100111010)

31.18.f = (1000000101111101)

31.19.f = (0100011000011111)

31.20.f = (0110110001101001)

31.21.f = (0100011111101000)

31.22.f = (1011100001100101)

31.23.f = (0011011000100111)

31.24.f = (1011001101001001)

31.25.f = (1110000110011100)

31.26.f = (1110110110000010)

31.27.f = (0001101010011110)

31.28.f = (0100111010011001)

31.29.f = (1110000101100011)

31.30.f = (1110001101100100)

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32.Заменить прочерки в векторе константами 0 и 1 так, чтобы получился вектор значений некоторой линейной функции.

32.1.= (1 1 00 )

32.2.= ( 010 1)

32.3.= (1 1 1 1)

32.4.= ( 1 1 0)

32.5.= (0 0 1 1 )

32.6.= ( 0 1 01 )

32.7.= ( 1 1 0 )

32.8.= ( 110 0 )

32.9.= ( 0 00 1 )

32.10.= ( 01 0 )

32.11.= ( 0 1 01)

32.12.= ( 1 0 0 0)

32.13.= ( 00 1 0 )

32.14.= ( 10 1 0 )

32.15.= ( 0 1 1 )

32.16.= ( 0 1 0)

32.17.= ( 00 0 1 )

32.18.= ( 1 0 0 1 )

32.19.= ( 0 1 01 )

32.20.= ( 1 0 1 )

32.21.= ( 0 0 1)

32.22.= ( 1 11 )

32.23.= ( 0 0 11 )

32.24.= (1 0 01 )

32.25.= (0 0 0 )

32.26.= ( 1 1 1 0 )

32.27.= ( 0 1 )

32.28.= ( 0 1 0 0 )

32.29.= ( 1 0)

32.30.= ( 00 1 1 )

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33.Используя лемму о нелинейной функции, подставèть на места переменных x1; x2; x3 функции из множества f0; 1; x; y; x; yg так, чтобы получи- лась конъюнкция xy или ее отрицание.

33.1.f(x1; x2; x3) = (11010101)

33.2.f(x1; x2; x3) = (11001000)

33.3.f(x1; x2; x3) = (01101111)

33.4.f(x1; x2; x3) = (01100010)

33.5.f(x1; x2; x3) = (01000101)

33.6.f(x1; x2; x3) = (00101110)

33.7.f(x1; x2; x3) = (11001101)

33.8.f(x1; x2; x3) = (00000110)

33.9.f(x1; x2; x3) = (10000110)

33.10.f(x1; x2; x3) = (10001000)

33.11.f(x1; x2; x3) = (11110101)

33.12.f(x1; x2; x3) = (10100110)

33.13.f(x1; x2; x3) = (10011011)

33.14.f(x1; x2; x3) = (11100011)

33.15.f(x1; x2; x3) = (10001101)

33.16.f(x1; x2; x3) = (10000011)

33.17.f(x1; x2; x3) = (01101101)

33.18.f(x1; x2; x3) = (11100010)

33.19.f(x1; x2; x3) = (01101010)

33.20.f(x1; x2; x3) = (01011000)

33.21.f(x1; x2; x3) = (00101111)

33.22.f(x1; x2; x3) = (00001001)

33.23.f(x1; x2; x3) = (11101010)

33.24.f(x1; x2; x3) = (00100111)

33.25.f(x1; x2; x3) = (01100101)

33.26.f(x1; x2; x3) = (01000110)

33.27.f(x1; x2; x3) = (11010111)

33.28.f(x1; x2; x3) = (00001101)

33.29.f(x1; x2; x3) = (11000100)

33.30.f(x1; x2; x3) = (10011110)

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34.Проверить, является ли функция монотонной.

34.1.(x1 _ x2) x1 x3

34.2.(x1 # x2) (x2 # x3)

34.3.x1x2(x1 ! x3)

34.4.(x1 # x3) _ (x1 j x2)

34.5.(x2 ! x3) (x1 j x3)

34.6.(x1 x1) j (x1 # x2)

34.7.x1 _ x3 _ (x1 j x2)

34.8.(x1 j x2) # x3

34.9.(x1 _ x2) ! (x2 x3)

34.10.(x3 ! x1) x1x2

34.11.(x1 x3) j (x1 # x2)

34.12.(x2 ! x1) _ (x1 # x3)

34.13.(x1 # x2) ! (x1 x3)

34.14.(x2 j x3) # (x1 x3)

34.15.(x1 ! x2) j (x2 x3)

34.16.(x1 x2) # (x1 _ x3)

34.17.(x1 j x2) x2x3

34.18.(x2 x3) (x1 _ x3)

34.19.(x1 _ x3) ! (x2 _ x3)

34.20.(x1 j x2) _ (x2 x1)

34.21.(x1 x3)(x3 ! x2)

34.22.(x1x3) j (x2 _ x3)

34.23.(x1 j x2) (x2 j x3)

34.24.(x2 _ x3) # (x3 ! x1)

34.25.(x1 x2)(x1 # x3)

34.26.(x2 _ x3) # (x1 # x3)

34.27.(x1 j x2) x2 x3

34.28.(x2 ! x1) # (x1 _ x3)

34.29.(x2 x3) # (x1 x2)

34.30.(x1 ! x3) j (x2 j x3)

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35.Проверить монотонность функции f, заданной векторно.

35.1.f = (0000011100110111)

35.2.f = (0000000100000111)

35.3.f = (0000011101110111)

35.4.f = (0001011100011111)

35.5.f = (0001010100011111)

35.6.f = (0000001100111111)

35.7.f = (0000000100111111)

35.8.f = (0000011100111111)

35.9.f = (0000001100011111)

35.10.f = (0000001101111111)

35.11.f = (0111011101110111)

35.12.f = (0000000100000011)

35.13.f = (0101010111111111)

35.14.f = (0000010100011111)

35.15.f = (0000010100010101)

35.16.f = (0101111101011111)

35.17.f = (0000000000010001)

35.18.f = (0000111100011111)

35.19.f = (0000000101011111)

35.20.f = (0000010101110111)

35.21.f = (0001010100111111)

35.22.f = (0000001101011111)

35.23.f = (0000000100010111)

35.24.f = (0000000101010111)

35.25.f = (0001001101110111)

35.26.f = (0000000100011111)

35.27.f = (0000011101011111)

35.28.f = (0011001100111111)

35.29.f = (0001000101010111)

35.30.f = (0001001100110011)

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